Question: You have found the following ages (in years) of all 5 lizards at your local zoo: $ 1,\enspace 2,\enspace 2,\enspace 1,\enspace 1$ What is the average age of the lizards at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 5 lizards at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{1 + 2 + 2 + 1 + 1}{{5}} = {1.4\text{ years old}} $ Find the squared deviations from the mean for each lizard. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $1$ year $-0.4$ years $0.16$ years $^2$ $2$ years $0.6$ years $0.36$ years $^2$ $2$ years $0.6$ years $0.36$ years $^2$ $1$ year $-0.4$ years $0.16$ years $^2$ $1$ year $-0.4$ years $0.16$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{0.16} + {0.36} + {0.36} + {0.16} + {0.16}} {{5}} $ $ {\sigma^2} = \dfrac{{1.2}}{{5}} = {0.24\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{0.24\text{ years}^2}} = {0.5\text{ years}} $ The average lizard at the zoo is 1.4 years old. There is a standard deviation of 0.5 years.